Question

Find the limit of x^x^2 as x approaches to 0.

Answer 1

\[\Large \lim_{x\to0}x^{(x^2)}\]

Let's use this familiar identity to rewrite our expression,\[\Large x^{(x^2)} \qquad=\qquad e^{\ln[x^{(x^2)}]}\]We can pass the limit into the exponential,\[\Large e^{\lim_{x\to0}\ln[x^{(x^2)}]}\]Ignoring the exponential base, let's solve the limit,\[\Large \lim_{x\to0}\ln[x^{(x^2)}] \qquad=\qquad \lim_{x\to0}(x^2)\ln x\]
From here we're getting the indeterminate form \[\Large 0\cdot-\infty\].
We want to get it into a form where we can apply L'Hopital's Rule.

This might seem a little tricky but we can rewrite our x^2 as,
\[\Large x^2=\frac{1}{\left(\frac{1}{x^2}\right)}\]
Which makes our limit become,
\[\Large lim_{x\to0}\frac{\ln x}{\frac{1}{x^2}}\]Now our limit is approaching the indeterminate form,
\[\Large \frac{-\infty}{\infty}\]
Yay! We can apply L'Hopital's! :)

Answer 2

Applying L'Hop gives us,

\[\Large lim_{x\to0} \frac{\left(\dfrac{1}{x}\right)}{\left(-2\dfrac{1}{x^3}\right)}\]Which we can write as,
\[\Large lim_{x\to0} \frac{\left(\dfrac{1}{x}\right)}{\left(-2\dfrac{1}{x^2}\right)\dfrac{1}{x}}\]
And then cancel some stuff,
\[\Large lim_{x\to0}\frac{1}{-2\dfrac{1}{x^2}}\]
Which we can rewrite as,
\[\Large lim_{x\to0}-\frac{1}{2}x^2\]

Answer 3

So now we can see that as x approaches 0, our function is also approaching zero.
BUT REMEMBER! This was all inside of our exponent!

\[\Large \lim_{x\to0}-\frac{1}{2}x^2=0\]

Answer 4

\[\Huge e^{\lim_{x\to0}\ln[x^{(x^2)}]} \qquad=\qquad e^0\]